Pythonic way to combine two lists in an alternating fashion?

Question

I have two lists, the first of which is guaranteed to contain exactly one more item than the second. I would like to know the most Pythonic way to create a

Answer 1

How about numpy? It works with strings as well:

import numpy as np

np.array([[a,b] for a,b in zip([1,2,3],[2,3,4,5,6])]).ravel()

Result:

array([1, 2, 2, 3, 3, 4])

Answer 2

I'm too old to be down with list comprehensions, so:

import operator
list3 = reduce(operator.add, zip(list1, list2))

Answer 3

This should do what you want:

>>> iters = [iter(list1), iter(list2)]
>>> print list(it.next() for it in itertools.cycle(iters))
['f', 'hello', 'o', 'world', 'o']

Answer 4

Without itertools and assuming l1 is 1 item longer than l2:

>>> sum(zip(l1, l2+[0]), ())[:-1]
('f', 'hello', 'o', 'world', 'o')

Using itertools and assuming that lists don't contain None:

>>> filter(None, sum(itertools.izip_longest(l1, l2), ()))
('f', 'hello', 'o', 'world', 'o')

Answer 5

I know the questions asks about two lists with one having one item more than the other, but I figured I would put this for others who may find this question.

Here is Duncan's solution adapted to work with two lists of different sizes.

list1 = ['f', 'o', 'o', 'b', 'a', 'r']
list2 = ['hello', 'world']
num = min(len(list1), len(list2))
result = [None]*(num*2)
result[::2] = list1[:num]
result[1::2] = list2[:num]
result.extend(list1[num:])
result.extend(list2[num:])
result

This outputs:

['f', 'hello', 'o', 'world', 'o', 'b', 'a', 'r'] 

Answer 6

Here's a one liner that does it:

list3 = [ item for pair in zip(list1, list2 + [0]) for item in pair][:-1]